A) 7
B) 4
C) 2
D) 1
Correct Answer: D
Solution :
Since,\[x=\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}=\sqrt{\frac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}}\] \[=\frac{2+\sqrt{3}}{\sqrt{4-3}}=2+\sqrt{3}\] \[\therefore \] \[{{x}^{2}}{{(x-4)}^{2}}={{(2+\sqrt{3})}^{2}}{{(2+\sqrt{3}-4)}^{2}}\] \[={{(\sqrt{3}+2)}^{2}}{{(\sqrt{3}-2)}^{2}}\] \[={{[(\sqrt{3})]}^{2}}-{{(2)}^{2}}{{]}^{2}}\] \[={{(3-4)}^{2}}={{(-1)}^{2}}\] \[=1\]
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